Magnetic field at the centre of a hydrogen-like atom (atomic number $= Z$) due to the motion of an electron in the $n^{th}$ orbit is proportional to

The de-Broglie wavelength of an electron moving in the $n^{\text{th}}$ Bohr orbit of radius $r$ is

According to Bohr's theory,the time-averaged magnetic field at the centre (i.e.,nucleus) of a hydrogen atom due to the motion of electrons in the $n^{th}$ orbit is proportional to ($n =$ principal quantum number).

In the lowest energy level of a hydrogen atom,the electron has an angular momentum of:

In a hydrogen atom,the electron and proton are bound at a distance of about $0.53 \; \mathring{A}$.
$(a)$ Estimate the potential energy of the system in $eV$,taking the zero of the potential energy at infinite separation of the electron from the proton.
$(b)$ What is the minimum work required to free the electron,given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in $(a)$?
$(c)$ What are the answers to $(a)$ and $(b)$ above if the zero of potential energy is taken at $1.06 \; \mathring{A}$ separation?

The radius of the innermost electron orbit of a hydrogen atom is $5.3 \times 10^{-11} \ m$. What is the radius of the $n=4$ state?